Problem: The lifespans of sloths in a particular zoo are normally distributed. The average sloth lives $14.1$ years; the standard deviation is $2.3$ years. Use the empirical rule (68-95-99.7%) to estimate the probability of a sloth living less than $11.8$ years.
Solution: $14.1$ $11.8$ $16.4$ $9.5$ $18.7$ $7.2$ $21$ $68\%$ $16\%$ $16\%$ We know the lifespans are normally distributed with an average lifespan of $14.1$ years. We know the standard deviation is $2.3$ years, so one standard deviation below the mean is $11.8$ years and one standard deviation above the mean is $16.4$ years. Two standard deviations below the mean is $9.5$ years and two standard deviations above the mean is $18.7$ years. Three standard deviations below the mean is $7.2$ years and three standard deviations above the mean is $21$ years. We are interested in the probability of a sloth living less than $11.8$ years. The empirical rule (or the 68-95-99.7 rule) tells us that $68\%$ of the sloths will have lifespans within 1 standard deviation of the average lifespan. The remaining $32\%$ of the sloths will have lifespans that fall outside the shaded area. Because the normal distribution is symmetrical, half $({16\%})$ will live less than $11.8$ years and the other half $({16\%})$ will live longer than $16.4$ years. The probability of a particular sloth living less than $11.8$ years is ${16\%}$.